POJ 2245 Lotto (枚举)

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本文介绍了一种针对德国乐透玩法的组合生成算法，该算法通过六重循环实现从指定集合中选择6个数字的所有可能组合。输入为大于6且小于13的整数k及相应数量的数字集合，输出为所有可能的游戏组合。

Lotto

http://poj.org/problem?id=2245

Time Limit: 1000MS

Memory Limit: 65536K

Description

In the German Lotto you have to select 6 numbers from the set {1,2,...,49}. A popular strategy to play Lotto - although it doesn't increase your chance of winning - is to select a subset S containing k (k > 6) of these 49 numbers, and then play several games with choosing numbers only from S. For example, for k=8 and S = {1,2,3,5,8,13,21,34} there are 28 possible games: [1,2,3,5,8,13], [1,2,3,5,8,21], [1,2,3,5,8,34], [1,2,3,5,13,21], ... [3,5,8,13,21,34].

Your job is to write a program that reads in the number k and the set S and then prints all possible games choosing numbers only from S.

Input

The input will contain one or more test cases. Each test case consists of one line containing several integers separated from each other by spaces. The first integer on the line will be the number k (6 < k < 13). Then k integers, specifying the set S, will follow in ascending order. Input will be terminated by a value of zero (0) for k.

Output

For each test case, print all possible games, each game on one line. The numbers of each game have to be sorted in ascending order and separated from each other by exactly one space. The games themselves have to be sorted lexicographically, that means sorted by the lowest number first, then by the second lowest and so on, as demonstrated in the sample output below. The test cases have to be separated from each other by exactly one blank line. Do not put a blank line after the last test case.

Sample Input

7 1 2 3 4 5 6 7
8 1 2 3 5 8 13 21 34
0

Sample Output

1 2 3 4 5 6
1 2 3 4 5 7
1 2 3 4 6 7
1 2 3 5 6 7
1 2 4 5 6 7
1 3 4 5 6 7
2 3 4 5 6 7

1 2 3 5 8 13
1 2 3 5 8 21
1 2 3 5 8 34
1 2 3 5 13 21
1 2 3 5 13 34
1 2 3 5 21 34
1 2 3 8 13 21
1 2 3 8 13 34
1 2 3 8 21 34
1 2 3 13 21 34
1 2 5 8 13 21
1 2 5 8 13 34
1 2 5 8 21 34
1 2 5 13 21 34
1 2 8 13 21 34
1 3 5 8 13 21
1 3 5 8 13 34
1 3 5 8 21 34
1 3 5 13 21 34
1 3 8 13 21 34
1 5 8 13 21 34
2 3 5 8 13 21
2 3 5 8 13 34
2 3 5 8 21 34
2 3 5 13 21 34
2 3 8 13 21 34
2 5 8 13 21 34
3 5 8 13 21 34

注意k<13，写一个六重循环没问题。

完整代码：

/*0ms,84KB*/

#include <cstdio>

int a[20];

int main(void)
{
	int n, i, j, k, p, q, r;
	while (scanf("%d", &n), n)
	{
		for (i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		for (i = 1; i <= n - 5; i++)
			for (j = i + 1; j <= n - 4; j++)
				for (k = j + 1; k <= n - 3; k++)
					for (p = k + 1; p <= n - 2; p++)
						for (q = p + 1; q <= n - 1; q++)
							for (r = q + 1; r <= n; r++)
								printf("%d %d %d %d %d %d\n", a[i], a[j], a[k], a[p], a[q], a[r]);
		printf("\n");
	}
	return 0;
}